Solution
= Solution
Expanding the <determinant> along its first row gives
$$
\det A=a(a^2-bc)-b(ac-b^2)+c(c^2-ab)
=a^3+b^3+c^3-3abc.
$$
The required factorization is
$$
\boxed{(a+b+c)(a^2+b^2+c^2-ab-bc-ca).}
$$
Each factor is a <symmetric polynomial>. The second is $\tfrac12[(a-b)^2+(b-c)^2+(c-a)^2]$, making its permutation symmetry especially transparent. This factorization holds over both real and complex coefficients.